Difference between revisions of "1995 AIME Problems/Problem 6"
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Thus, our general formula for <math>n=p_1^{k_1}p_2^{k_2}</math> is | Thus, our general formula for <math>n=p_1^{k_1}p_2^{k_2}</math> is | ||
| − | + | Number of factors that satisfy the above <math>=(2k_1k_2+k_1+k_2)-(k_1k_2+k_1+k_2)=k_1k_2</math> | |
| − | Incorporating this into our problem gives < | + | Incorporating this into our problem gives <math>19\times31=\boxed{589}</math>. |
== See also == | == See also == | ||
Revision as of 15:27, 16 September 2014
Contents
Problem
Let 
How many positive integer divisors of 
are less than 
but do not divide ?
Solution 1
We know that 
must have 
factors by its prime factorization. If we group all of these factors (excluding 
) into pairs that multiply to , then one factor per pair is less than , and so there are 
factors of that are less than . There are 
factors of , which clearly are less than , but are still factors of . Therefore, there are 
factors of that do not divide .
Solution 2
Let 
for some prime 
. Then has 
factors less than .
This simplifies to 
.
The number of factors of less than is equal to 
.
Thus, our general formula for is
Number of factors that satisfy the above 
Incorporating this into our problem gives 
.
See also
| 1995 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
